What is the integral of cot^5(2x)csc^4(2x) ?

The integral of cot^5(2x)csc^4(2x) is -1/2 (1/6 cot^6(2x)+1/8 cot^8(2x))+C

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integral of cot^5(2x)csc^4(2x)

\int cot^{5} (2 x) csc^{4} (2 x) d x

- \frac{1}{2} (\frac{1}{6} cot^{6} (2 x) + \frac{1}{8} cot^{8} (2 x)) + C

Solution steps

\int cot^{5} (2 x) csc^{4} (2 x) d x

Apply u-substitution

= \int cot^{5} (u) csc^{4} (u) \frac{1}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int cot^{5} (u) csc^{4} (u) d u

Apply u-substitution

= \frac{1}{2} \cdot \int - v^{5} (1 + v^{2}) d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} (- \int v^{5} (1 + v^{2}) d v)

Expand

v^{5} (1 + v^{2}) : v^{5} + v^{7}

= \frac{1}{2} (- \int v^{5} + v^{7} d v)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- (\int v^{5} d v + \int v^{7} d v))

\int v^{5} d v = \frac{v ^{6}}{6}

\int v^{7} d v = \frac{v ^{8}}{8}

= \frac{1}{2} (- (\frac{v ^{6}}{6} + \frac{v ^{8}}{8}))

Substitute back

= \frac{1}{2} (- (\frac{cot ^{6} ( 2 x )}{6} + \frac{cot ^{8} ( 2 x )}{8}))

Simplify

\frac{1}{2} (- (\frac{cot ^{6} ( 2 x )}{6} + \frac{cot ^{8} ( 2 x )}{8})) : - \frac{1}{2} (\frac{1}{6} cot^{6} (2 x) + \frac{1}{8} cot^{8} (2 x))

= - \frac{1}{2} (\frac{1}{6} cot^{6} (2 x) + \frac{1}{8} cot^{8} (2 x))

Add a constant to the solution

= - \frac{1}{2} (\frac{1}{6} cot^{6} (2 x) + \frac{1}{8} cot^{8} (2 x)) + C

What is the integral of cot^5(2x)csc^4(2x) ?
The integral of cot^5(2x)csc^4(2x) is -1/2 (1/6 cot^6(2x)+1/8 cot^8(2x))+C